80分,2TLE,求调

P1440 求m区间内的最小值

ETO_ @ 2024-08-11 15:15:36

#include<bits/stdc++.h>
using namespace std;
int n,m;
int a[2000006];
int ans[2000006];
int main(){
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        ans[i]=a[i];
    }
    for(int j=1;(1<<j)<=m;j++){
        for(int i=1;i+(1<<j)<=n;i++){
            ans[i]=min(ans[i],ans[i+(1<<(j-1))]);
        }
    }
    cout<<0<<endl;
    int l,r,t,mi=INT_MAX;
    for(int i=2;i<=m;i++){
        mi=min(mi,a[i-1]);
        cout<<mi<<endl;
    }
    for(int i=m+1;i<=n;i++){
        l=i-m;
        r=i-1;
        t=log2(r-l+1);
        cout<<min(ans[l],ans[r-(1<<t)+1])<<endl;
    }
    return 0;
} 

by SX114514 @ 2024-08-11 15:21:20

用scanf!!!!!!!!!!!!!!!!!


by Scez @ 2024-08-13 07:58:52

@ETO_ 输出用printf


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